A short question concerning the distributional solution of $xf=0$

Question

I was reading my notes on the following result:

All the $\mathcal{D}'(\mathbb{R})$ solutions to $xf =0$ are of the form $c\delta $ where $c$ is constant and $\delta$ is the dirac delta distribution

The sketch of the proof goes like:

For test function $\phi(x)$, write it like $(\phi(x)-\phi(0)+\phi(0))\phi_1(x)$ where $\phi_1$ is a test function equals $1$ on support of $\phi$, then $\langle f,\phi\rangle=\langle f,(\phi(x)-\phi(0)\phi_1)\rangle+\langle f,\phi(0) \phi_1\rangle = \phi(0)\langle f,\phi_1\rangle$.

It was mentioned in class that $\phi_1$ is dependent on $\phi$, but it is possible to fix a single $\phi_1$ which works for all $\phi$, however I was stuck on this task, so I decided to ask here, any hints would be appreciated.

Answer 1

I think what is meant is: if $\phi_1(x)$ and $\phi_2(x)$ are any test functions that are both $1$ in a neighbourhood of $0$, then $\langle f, \phi_1(x) \rangle = \langle f, \phi_2(x)\rangle$ (because $\phi_1(x) - \phi_2(x) = \psi(x) x$ for a test function $\psi(x)$), so it doesn't matter which $\phi_1$ you take.

Answer 2

Another approach is to take $\theta(x)$ - a test function which is $1$ in a neighbourhood of zero. Then you study $\phi(x) -\phi(0 )\theta(x)$ and arrive to the same conclusion.